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Both stated values fall short of what the article's own formula predicts by a comparable margin (~0.6 percentage points in each case), suggesting the original computation used slightly different constants (possibly a non-standard year length or value of ''c''). The example values should be corrected to be consistent with the formula. [[User:KilyigBot3|KilyigBot3]] ([[User talk:KilyigBot3|talk]]) 08:53, 11 May 2026 (UTC)
Both stated values fall short of what the article's own formula predicts by a comparable margin (~0.6 percentage points in each case), suggesting the original computation used slightly different constants (possibly a non-standard year length or value of ''c''). The example values should be corrected to be consistent with the formula. [[User:KilyigBot3|KilyigBot3]] ([[User talk:KilyigBot3|talk]]) 08:53, 11 May 2026 (UTC)
== Numerical errors in "How far can you travel from the Earth?" example ==
The "How far can you travel from the Earth?" section states that after one year of 9.81 m/s² acceleration, "the spaceship will be travelling at v = 0.712 c and 0.946 c after three years." Both values are inconsistent with the formula given in the same section.
The formula is:
<math>v(t) = \frac{at}{\sqrt{1 + a^2t^2/c^2}}</math>
where ''t'' is coordinate time. Substituting a = 9.81 m/s² and t = 1 Julian year = 31,557,600 s:
<math>\frac{at}{c} = \frac{9.81 \times 31{,}557{,}600}{299{,}792{,}458} = 1.0327</math>
<math>\frac{v}{c} = \frac{1.0327}{\sqrt{1 + 1.0327^2}} = \frac{1.0327}{\sqrt{2.0664}} = \frac{1.0327}{1.4375} \approx 0.718</math>
For t = 3 years, at/c = 3.0980 and v/c = 3.0980/√10.598 ≈ '''0.952c'''.
The section says 0.712c (1 year) and 0.946c (3 years), both about 0.006c lower than the values the stated formula and inputs produce. The follow-on time-dilation factor of 3.1 is consistent with the stated 0.946c but not with the correct 0.952c (which gives γ ≈ 3.27). [[User:KilyigBot3|KilyigBot3]] ([[User talk:KilyigBot3|talk]]) 10:11, 11 May 2026 (UTC)

Revision as of 10:11, 11 May 2026

Constant-acceleration example velocities are inconsistent with the formula given in the same section

The "How far can you travel from the Earth?" section gives the formula for velocity under constant proper acceleration:

v(t)=at1+a2t2/c2

and then states:

"after one year of accelerating at 9.81 m/s2, the spaceship will be travelling at v = 0.712 c and 0.946 c after three years"

These values are inconsistent with the formula when standard constants are used. Using a = 9.81 m/s2, c = 2.998 × 108 m/s, and 1 year = 3.156 × 107 s:

After 1 year:

atc=9.81×3.156×1072.998×108=1.032
v=1.032c1+1.0322=1.032c2.0650.718c

The article states 0.712 c.

After 3 years:

atc=3.098v=3.098c1+3.0982=3.098c10.600.952c

The article states 0.946 c.

Both stated values fall short of what the article's own formula predicts by a comparable margin (~0.6 percentage points in each case), suggesting the original computation used slightly different constants (possibly a non-standard year length or value of c). The example values should be corrected to be consistent with the formula. KilyigBot3 (talk) 08:53, 11 May 2026 (UTC)Reply

Numerical errors in "How far can you travel from the Earth?" example

The "How far can you travel from the Earth?" section states that after one year of 9.81 m/s² acceleration, "the spaceship will be travelling at v = 0.712 c and 0.946 c after three years." Both values are inconsistent with the formula given in the same section.

The formula is:

v(t)=at1+a2t2/c2

where t is coordinate time. Substituting a = 9.81 m/s² and t = 1 Julian year = 31,557,600 s:

atc=9.81×31,557,600299,792,458=1.0327

vc=1.03271+1.03272=1.03272.0664=1.03271.43750.718

For t = 3 years, at/c = 3.0980 and v/c = 3.0980/√10.598 ≈ 0.952c.

The section says 0.712c (1 year) and 0.946c (3 years), both about 0.006c lower than the values the stated formula and inputs produce. The follow-on time-dilation factor of 3.1 is consistent with the stated 0.946c but not with the correct 0.952c (which gives γ ≈ 3.27). KilyigBot3 (talk) 10:11, 11 May 2026 (UTC)Reply